Linear Spaces

Basis

Definition: Let VV be a vector space. A (finite) set of vectors S={v1,v2,…,vn}S=\{v_1, v_2, \dots, v_n\} is called a basis set for VV iff

A (finite dimensional) linear space VV has many bases. All bases of a linear space have the same number of elements. This number is called the dimension of the linear space.


Example: V=R2V = \mathbb{R}^2, consider the two bases:

S1={[01],[10]}S_1 = \begin{Bmatrix}\begin{bmatrix} 0 \\ 1 \end{bmatrix},\begin{bmatrix} 1 \\ 0 \end{bmatrix}\end{Bmatrix}

  1. S1S_1 is linearly independent set since c1[01]+c2[10]=[00]c_1\begin{bmatrix} 0 \\ 1 \end{bmatrix} + c_2\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} implies c1=c2=0c_1 = c_2 = 0, ∀c1,c2∈R(F)\forall c_1 , c_2 \in \mathbb{R}(F).
  2. Span(S1)=V=R2Span(S_1) = V = \mathbb{R}^2 Hence S1S_1 is a basis for VV.

S2={[11],[23]}S_2 = \begin{Bmatrix}\begin{bmatrix} 1 \\ 1 \end{bmatrix},\begin{bmatrix} 2 \\ 3 \end{bmatrix}\end{Bmatrix}

  1. S2S_2 is linearly independent set since c1[11]+c2[23]=[00]c_1\begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2\begin{bmatrix} 2 \\ 3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} implies c1=c2=0c_1 = c_2 = 0, ∀c1,c2∈R(F)\forall c_1 , c_2 \in \mathbb{R}(F).
  2. Span(S2)=V=R2Span(S_2) = V = \mathbb{R}^2

Example: V=R2V = \mathbb{R}^2, and F=RF = \mathbb{R}, consider the base:
B={[10],[11]}B = \begin{Bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix},\begin{bmatrix} 1 \\ 1 \end{bmatrix}\end{Bmatrix} y=[23]y = \begin{bmatrix} 2 \\ 3 \end{bmatrix} [y]B[y]_B=?

Solution:

  1. BB is linearly independent set since c1[10]+c2[11]=[00]c_1\begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} implies c1=c2=0c_1 = c_2 = 0, ∀c1,c2∈R(F)\forall c_1 , c_2 \in \mathbb{R}(F).
  2. Span(B)=V=R2Span(B) = V = \mathbb{R}^2
  3. [y]B=[c1c2][y]_B = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}
  4. y=c1[10]+c2[11]=[c1+c2c2]y = c_1\begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} c_1 + c_2 \\ c_2 \end{bmatrix}
  5. c1+c2=2c_1 + c_2 = 2 and c2=3c_2 = 3
  6. c1=−1c_1 = -1 and c2=3c_2 = 3
  7. [y]B=[−13][y]_B = \begin{bmatrix} -1 \\ 3 \end{bmatrix}

Example: V=Span{cos(t),sin(t)}V = Span\{cos(t), sin(t)\}, and F=RF = \mathbb{R}, consider the base:
B={cos(t),sin(t)}B = \begin{Bmatrix}cos(t), sin(t)\end{Bmatrix} y=cos(t−π3)y = cos(t-\frac{\pi}{3}) [y]B[y]_B=?

Solution:

  1. BB is linearly independent set since c1cos(t)+c2sin(t)=0c_1cos(t) + c_2sin(t) = 0 implies c1=c2=0c_1 = c_2 = 0, ∀c1,c2∈R(F)\forall c_1 , c_2 \in \mathbb{R}(F).
  2. Span(B)=V=Span{cos(t),sin(t)}Span(B) = V = Span\{cos(t), sin(t)\}
  3. [y]B=[c1c2][y]_B = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}
  4. y=c1cos(t)+c2sin(t)=cos(t−π3)=cos(t)cos(π3)+sin(t)sin(π3)=12cos(t)+32sin(t)y = c_1cos(t) + c_2sin(t) = cos(t-\frac{\pi}{3}) = cos(t)cos(\frac{\pi}{3}) + sin(t)sin(\frac{\pi}{3}) = \frac{1}{2}cos(t) + \frac{\sqrt{3}}{2}sin(t)
  5. c1=12c_1 = \frac{1}{2} and c2=32c_2 = \frac{\sqrt{3}}{2}
  6. [y]B=[1232][y]_B = \begin{bmatrix} \frac{1}{2} \\ \frac{\sqrt{3}}{2} \end{bmatrix}

Claim: For a given basis BB, the representation [y]B[y]_B of a vector yy is unique.
Proof: By contradiction.
Assume [y]B=[c1c2][y]_B = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} and [y]B=[d1d2][y]_B = \begin{bmatrix} d_1 \\ d_2 \end{bmatrix}
Then y=c1b1+c2b2=d1b1+d2b2y = c_1b_1 + c_2b_2 = d_1b_1 + d_2b_2
c1b1+c2b2−d1b1−d2b2=0c_1b_1 + c_2b_2 - d_1b_1 - d_2b_2 = 0
(c1−d1)b1+(c2−d2)b2=0(c_1 - d_1)b_1 + (c_2 - d_2)b_2 = 0
Since BB is linearly independent, c1−d1=0c_1 - d_1 = 0 and c2−d2=0c_2 - d_2 = 0
Hence c1=d1c_1 = d_1 and c2=d2c_2 = d_2 â– \blacksquare

Remark: The representation of a vector yy in a basis BB is unique. The representation of a vector yy in a basis BB is called the coordinate vector of yy with respect to BB.


Ordered Basis

Definition: Let VV be a vector space. An ordered set of basis vectors S={v1,v2,…,vn}S=\{v_1, v_2, \dots, v_n\} is called an ordered basis for VV. If y=(x1,x2,...,xn)y = (x_1,x_2,...,x_n) is an ordered basis for VV, then every vector x∈Vx \in V can be written as a linear combination of the basis vectors as follows:

Theorem: Let VV be an n-dimensional vector space over R\mathbb{R}. Let B1B_1 and B2B_2 be two bases for VV. Then there exists a unique n×nn \times n real invertible matrix PP such that [x]B2=P[x]B1[x]_{B_2} = P[x]_{B_1} for all x∈Vx \in V.

Proof: By construction.

Example: Consider V=V = polynomials of degree ≤2\leq 2 with coefficients in R\mathbb{R} and the bases:
B1={1,t−1,(t−1)2}B_1 = \begin{Bmatrix}1, t-1, (t-1)^2\end{Bmatrix}
B2={1,t,t2}B_2 = \begin{Bmatrix}1, t, t^2\end{Bmatrix}
Find the matrix PP such that [x]B1=P[x]B2[x]_{B_1} = P[x]_{B_2} for all x∈Vx \in V.

Solution:

  1. [x]B1=[c1c2c3][x]_{B_1} = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} and [x]B2=[d1d2d3][x]_{B_2} = \begin{bmatrix} d_1 \\ d_2 \\ d_3 \end{bmatrix}
  2. x=c1+c2(t−1)+c3(t−1)2=c1+c2t−c2+c3t2−2c3t+c3x = c_1 + c_2(t-1) + c_3(t-1)^2 = c_1 + c_2t - c_2 + c_3t^2 - 2c_3t + c_3
  3. x=(c1−c2+c3)+(c2−2c3)t+c3t2x = (c_1 - c_2 + c_3) + (c_2 - 2c_3)t + c_3t^2
  4. d1=c1−c2+c3d_1 = c_1 - c_2 + c_3
  5. d2=c2−2c3d_2 = c_2 - 2c_3
  6. d3=c3d_3 = c_3
  7. c1=d1+d2+d3c_1 = d_1 + d_2 + d_3
  8. c2=d2+2d3c_2 = d_2 + 2d_3
  9. c3=d3c_3 = d_3
  10. [x]B1=[d1+d2+d3d2+2d3d3]=[111012001][d1d2d3]=P[x]B2[x]_{B_1} = \begin{bmatrix} d_1 + d_2 + d_3 \\ d_2 + 2d_3 \\ d_3 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} d_1 \\ d_2 \\ d_3 \end{bmatrix} = P[x]_{B_2}
  11. P=[111012001]P = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix}

Example: Show that Matrix PP is invertible.
 ~Proof: Let VV be an n-dimensional vector space over Rn×n\mathbb{R^{n\times n}}. B1B_1 and B2B_2 are bases for VV. A vector vv holds,


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